Question One

It is 2011 and the Minnesota State Fair is just around the corner. Ryan can’t wait to go and stand in line for all of his favorite state fair foods. He always make sure that he stops at the following booths: Deep Fried Candy Bar $3.50, Cotton Candy $3.25, Corn on the Cob $4.00, Pronto Pup $3.75, and the Pork Chop on a Stick $5.25. Ryan heard from Lena that all the food booths at the state fair are increasing their prices by 7%. If this is the case how much will he now have to pay for all of his favorite foods?

Helpful Advice

For this problem using scalar multiplication

would be a very useful tool.

Step 1: Make a Matrix

2010Prices

Deep Fried Candy Bar 3.50Cotton Candy 3.25Corn on the Cob 4.00Pronto Pups 3.75Pork Chop on a Stick 5.25

Scalar * Matrix 1 + Matrix 1 Setting up the matrix problem

Scalar *

a

=

Scalar * ab Scalar * bc Scalar * cd Scalar * de Scalar * e

Scalar Multiplication

Review

Scalar Addition Review

a

+

f

=

a + fb g b + gc h c + hd i d + ie j e + j

Run the numbers

0.08 *

3.50

+

3.503.25 3.254.00 4.003.75 3.755.25 5.25

0.28

+

3.50

=

3.780.26 3.25 3.510.32 4.00 4.320.30 3.75 4.050.42 5.25 5.67

Another Route 1.08 *

3.503.254.003.755.25

By having your scalar equal to 1.08 you are eliminating the need to add the percentage increase to the original amount – this new scalar will do just that!

Just multiple the matrix by the scalar and you will get your result immediately!

=

3.783.514.324.055.67

Final Solution

2010Prices

2011 Price

sDeep Fried Candy Bar 3.50 3.78Cotton Candy 3.25 3.51Corn on the Cob 4.00 4.32Pronto Pups 3.75 4.05Pork Chop on a Stick 5.25 5.67

Question Two

Solve the following system of linear equations.

X + 4Y = 32X + 9Y = 5

NotationRow switching : A row within the matrix can

be switched with another row. R1 R2 Row multiplication: Each element in a row can be multiplied by a non-zero constant.k* R1 R1, k ≠ 0Row addition: A row can be replaced by the sum of that row and a multiple of another row. R1 + kR2 R1, k ≠ 0

Goal

1 0 0 #1

0 1 0 #2

The ultimate goal is to get our matrix to look like the one to the left and then the values in the #1, #2 locations will be the solutions to the problem for x and y

Using row operations here we go….Step 1: Set up the augmented matrix

1 4 32 9 5

Step 2: First Operation

Step 2: First Operation

-2R1 + R2 R2

1 4 30 1 -1

Notice no change was made to R1

(row 1)

R1 = Row One and R2 = Row Two

Step 3: Second Operation

-4R2 + R1 R1

1 0 70 1 -1 Notice no

change was made to R2

(row 2)

R1 = Row One and R2 = Row Two

Step 4: Finalize Solution

1 0 70 1 -1

X Y

X = 7 Y = -1

Step 5: Double check the solution you found

X + 4Y = 32X + 9Y = 5

Original Equations

(7) + 4(-1) = 32(7) + 9(-1) = 5

It Works!

Question Three

Solve the following system of linear equations.

X + Y - Z = -22X – Y + Z= 5-X+2Y+2Z = 1

Same idea but now we have to work with the three variables.

Goal

The ultimate goal is to get our matrix to look like the one to the left and then the values in the #1, #2, #3 locations will be the solutions to the problem for x and y and z

1 0 0 #10 1 0 #20 0 1 #3

Using row operations here we go….Step 1: Set up the augmented matrix

1 1 -1 -22 -1 1 5-1 2 2 1

Step 2: First Operation

-2R1 + R2 R2

Notice no change was made to R1

and R3

R1 = Row One and R2 = Row Two and R3 = Row Three

1 1 -1 -20 -3 3 9-1 2 2 1

Step 3: Operation

R1 + R3 R3

Notice no change was made to R1

and R2

R1 = Row One and R2 = Row Two and R3 = Row Three

1 1 -1 -20 -3 3 90 3 1 -1

Step 4: Operation

R2 ÷ -3 R2

Notice no change was made to R1

and R3

R1 = Row One and R2 = Row Two and R3 = Row Three

1 1 -1 -20 1 -1 -30 3 1 -1

Step 5: Operation

-3R2 + R3 R3

Notice no change was made to R1

and R2

R1 = Row One and R2 = Row Two and R3 = Row Three

1 1 -1 -20 1 -1 -30 0 4 8

Step 6: Operation

R3 ÷ 4 R3

Notice no change was made to R1

and R2

R1 = Row One and R2 = Row Two and R3 = Row Three

1 1 -1 -20 1 -1 -30 0 1 2

Step 7: Finalize Solution

1 1 -1 -20 1 -1 -30 0 1 2

X + Y - Z = -2Y - Z= -3

Z = 2

Now:

Then: Y - Z = -3Y - 2= -3

Y = -1

Lastly: X + Y - Z = -2X + (-1) – 2 = -2

X-3 = -2X= 1

Step 8: Double check the solution you found

X + Y - Z = -22X – Y + Z= 5-X+2Y+2Z = 1

Original Equations

(1) + (-1) – (2) = -22(1) – (-1) + (2)= 5-(1)+2(-1)+2(2) = 1

It Works!!

Question Four

Now to wrap it all up we want to determine how much it actually costs to get into the fair and enjoy this fun get together! As you saw on the clip, one day at the fair just isn’t enough!! At the gates on opening day they are selling one day passes as well as bonus three day passes. We were only able to collect the data for 2 particular gates on this day. Gate A sells 2,090 day passes and 980 three day passes. Gate B sells 1,807 day passes and 712 three day passes. If Gate A collected $31,290 and Gate B collected $24,753, how much does it cost to by a one day pass and a three day pass?

Helpful Advice

For this problem finding the determinate and

the inverse would be a very useful tool.

Set up a Matrix Matrix 1: A B

C DMatrix Format

1-day 3-day Total $Gate A 2090 980 $31,290Gate B 1807 712 $24,753

Matrix specific to this example

Finding the determinantDeterminant = A*D - B*C

2090 9801807 712

A BC D

2090 * 712 - 980 * 1807

Det A = -282,780

Finding the inverse of a matrix Inverse Matrix = 1 * D -B

Det A -C A

-3.5363E-06 * 712 -980 = -0.0025 0.0035

-1807 2090 0.0064 -0.0074

Multiplying inverse and solution to find unknown

A B x E

C D F

= (A * E) + (B * F)

(C * E) + (D * F)

-0.002518 0.003466 X 31290

0.006390 -0.007391 24753

= $7

$17

Double check the solutions to make sure they do indeed work in the problem

2090 980 X $7

1807 712 $17

= $31,290$24,753

Final Solution

One Day Pass $7.00Three Day Pass $17.00

In the future you can go to this Excel document to plug in your matrix information and have it automatically calculate